Problem: Complete the square to solve for $x$. $x^{2}+20x+99 = 0$
Begin by moving the constant term to the right side of the equation. $x^2 + 20x = -99$ We complete the square by taking half of the coefficient of our $x$ term, squaring it, and adding it to both sides of the equation. Since the coefficient of our $x$ term is $20$ , half of it would be $10$ , and squaring it gives us ${100}$ $x^2 + 20x { + 100} = -99 { + 100}$ We can now rewrite the left side of the equation as a squared term. $( x + 10 )^2 = 1$ Take the square root of both sides. $x + 10 = \pm1$ Isolate $x$ to find the solution(s). $x = -10\pm1$ So the solutions are: $x = -9 \text{ or } x = -11$ We already found the completed square: $( x + 10 )^2 = 1$